Question

Calculate $\mathbf{∆}{\mathbf{S}}_{\mathbf{surr}}$for the following reactions at 25°C and1 atm.

1. ${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{8}}\mathbf{}\left(g\right)\mathbf{}\mathbf{+}\mathbf{5}{\mathbf{O}}_{\mathbf{2}}\mathbf{}\left(g\right)\mathbf{}\mathbf{\to }\mathbf{3}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{4}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\left(I\right)\mathbf{}\mathbf{∆}{\mathbf{H}}^{\mathbf{°}}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{2221}\mathbf{kJ}$
2. $\mathbf{2}{\mathbf{NO}}_{\mathbf{2}}\mathbf{}\left(g\right)\mathbf{}\mathbf{\to }\mathbf{2}\mathbf{NO}\mathbf{}\left(g\right)\mathbf{}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{}\left(g\right)\mathbf{}\mathbf{∆}{\mathbf{H}}^{\mathbf{°}}\mathbf{=}\mathbf{112}\mathbf{kJ}$

Short answer

Answer

$\mathbf{∆}{\mathbf{S}}_{\mathbf{surr}}$ for the following reactions at 25°C and 1 atm are,

1. ${\mathrm{C}}_3{\mathrm{H}}_8\left(\mathrm{g}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)\to 3{\mathrm{CO}}_2\left(\mathrm{g}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right)∆{\mathrm{H}}^{°}=-2221\mathrm{kJ}\mathrm{is}7.45\times {10}^3\mathrm{J}/\mathrm{K}$
   
2. $2{\mathrm{NO}}_2\left(\mathrm{g}\right)\to 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)∆{\mathrm{H}}^{°}=112\mathrm{kJ}\mathrm{is}\mathbf{-}\mathbf{376}\mathbf{J}\mathbf{/}\mathbf{K}$

Step-by-step solution

Step 1: Introduction to the Concept

Entropy is a characteristic of a system that quantifies the unpredictability or degree of chaos. It's a state function, after all. With an increase in the universe's entropy,spontaneous change occurs.

$\mathbf{∆}{\mathbf{S}}_{\mathbf{univ}}\mathbf{=}\mathbf{∆}{\mathbf{S}}_{\mathbf{sys}}\mathbf{+}\mathbf{∆}{\mathbf{S}}_{\mathbf{surr}}-----\left(1\right)$

The entropy change of the systemand its surroundings equal to the entropy change of the cosmos.

The following is the relationship between entropy change and enthalpy change for the surrounding:

$\begin{array}{rcl}\mathbf{∆}{\mathbf{S}}_{\mathbf{surr}}& \mathbf{=}& {}_{\mathbf{-}}\mathbf{∆}\mathbf{H}\mathbf{/}\mathbf{T}-----\left(2\right)\\ ∆{\mathrm{S}}_{\mathrm{surr}}& =& \mathrm{Entropy}\mathrm{change}\mathrm{of}\mathrm{the}\mathrm{environment}\\ ∆\mathrm{H}& =& \mathrm{Enthalpy}\mathrm{change}\\ \mathrm{T}& =& \mathrm{Temperature}\end{array}$

Step 2: Solution Explanation

a)

From the given,

${\mathrm{C}}_3{\mathrm{H}}_8\left(\mathrm{g}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)\to 3{\mathrm{CO}}_2\left(\mathrm{g}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right)∆{\mathrm{H}}^{°}=-2221\mathrm{kJ}$

From the equation (2),

$\begin{array}{rcl}∆{\mathrm{S}}_{\mathrm{surr}}& =& -\left(-2221\mathrm{kJ}\right)/298\mathrm{K}\\ & =& +7.45\mathrm{kJ}/\mathrm{K}\\ & =& 7.45\times {10}^3\mathrm{J}/\mathrm{K}\end{array}$

Therefore, $\mathbf{∆}{\mathbf{S}}_{\mathbf{surr}\mathbf{}\mathbf{}}$is $\mathbf{7}\mathbf{.}\mathbf{45}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{K}$

Step 3: Solution Explanation

b)

From the given,

$2{\mathrm{NO}}_2\left(\mathrm{g}\right)\to 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)∆{\mathrm{H}}^{°}=112\mathrm{kJ}$

From the equation (2),

$\begin{array}{rcl}∆{\mathrm{S}}_{\mathrm{surr}}& =& -\left(+112\mathrm{kJ}\right)\times 298\mathrm{K}\\ & =& -0.376\mathrm{kJ}/\mathrm{K}\\ & =& -376\mathrm{J}/\mathrm{K}\end{array}$

Therefore, $\mathbf{∆}{\mathbf{S}}_{\mathbf{surr}}$is -376J/K